poj 2325 Persistent Numbers （贪心+高精度）

把输入数字每次从9-2除，能整除则记录该数字，最后从小到大输出。

应该算是水题，不过窝第一次写高精度除法，虽然1A，不过中间改了好多次。

/******************************************
Problem: 2325		User:
Memory: 684K		Time: 110MS
Language: G++		Result: Accepted
******************************************/
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char numStr[1005];
int num[1005];
int num1[1005];
int ans[1005];
int len;
int cnt;
int in;//因为数字位数会越除越小，用in来记录少了多少位

bool div(int n)
{
	int i;
	int temp = (num[in] < n);
	for (i = in; i < len; ++i) num1[i] = num[i];
	for (i = in; i < len; ++i) {
		num1[i + 1] += (num1[i] % n) * 10;
		num1[i] /= n;
	}
	if (num1[i]) {
		num1[i] = 0;
		return false;
	} else {
		in += temp;
		for (i = in; i < len; ++i) num[i] = num1[i];
		return true;
	}
}

int main()
{
	int i;
    while (scanf("%s", numStr) != EOF) {
		len = strlen(numStr);
		if (len == 2 && numStr[0] == '-' && numStr[1] == '1')
			break;
		if (len == 1) {
			printf("%c%s\n", '1', numStr);
			continue;
		}
		in = cnt = 0;
		memset(num1, 0, sizeof num1);
		memset(num, 0, sizeof num);
		for (i = 0; i < len; ++i) {
			num[i] = numStr[i] - '0';
		}
		while (!(len - in == 1 && num[in] == 1)) {
			//printf("in:%d\n", in);
			for (i = 9; i >= 2; --i) {
				if (div(i)) {
					ans[cnt++] = i;
					break;
				}
			}
			//printf("i=%d\n", i);
			if (i == 1) break;
		}
		if (i == 1) printf("There is no such number.");
		else
			for (i = cnt - 1; i >= 0; --i)
				printf("%d", ans[i]);
		printf("\n");
    }
    return 0;
}